Question

$\underset{x\to -1}{lim}\frac{1}{\sqrt{|x|-\{-x\}}}$ (where {x} denotes the fractional part of x) is equal to

• A. Does not Exist
• B. $1$
• C. $\infty$
• D. $\frac{1}{2}$

Answer 1

The correct option is A Does not Exist

$L.H.L=\underset{x\to {-1}^{-}}{lim}\frac{1}{\sqrt{|x|-\{-x\}}}=\underset{x\to {-1}^{-}}{lim}\frac{1}{\sqrt{-x-(x+2)}}$$=\underset{x\to {-1}^{-}}{lim}\frac{1}{\sqrt{-2x-2}}=\infty$
$R.H.L.=\underset{x\to {-1}^{+}}{lim}\frac{1}{\sqrt{|x|-\{-x\}}}=\frac{1}{\sqrt{-x-(x+1)}}$
$=\underset{x\to {-1}^{+}}{lim}\frac{1}{\sqrt{-2x-1}}=1$
$\text{Hence, the limit does not exist.}$